Find the arc length of the curve x = 1/2 ((y²)/2 - ln y)

Find the arc length of the curve $x = \dfrac{1}{2}\left(\dfrac{y^2}{2} - \ln y\right)$ between the points with ordinates $y = 1$ and $y = e$.

We treat $x$ as a function of $y$ and use the arc-length formula $L=\int\sqrt{1+(dx/dy)^2}\,dy$.

Step 1: Differentiate

\[x=\frac{y^2}4-\frac{\ln y}2 \implies \frac{dx}{dy}=\frac y2-\frac1{2y}=\frac{y^2-1}{2y}\]

Step 2: Simplify $1+(dx/dy)^2$

\[1+\left(\frac{y^2-1}{2y}\right)^2 = \frac{4y^2+(y^2-1)^2}{4y^2}=\frac{y^4+2y^2+1}{4y^2}=\frac{(y^2+1)^2}{4y^2}\]

For $y\in[1,e]$ (positive), taking the square root:

\[\sqrt{1+(dx/dy)^2}=\frac{y^2+1}{2y}=\frac12\left(y+\frac1y\right)\]

Step 3: Integrate

\[L=\frac12\int_1^e\left(y+\frac1y\right)dy = \frac12\left[\frac{y^2}2+\ln y\right]_1^e\]

Step 4: Evaluate

At $y=e$: $\frac{e^2}2+1$. At $y=1$: $\frac12$.

\[L=\frac12\left[\left(\frac{e^2}2+1\right)-\frac12\right]=\frac12\cdot\frac{e^2+1}2=\frac{e^2+1}4\]

\[\boxed{L=\frac{e^2+1}4\approx2.097}\]